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# Sums of independent random variables

This lecture discusses how to derive the distribution of the sum of two independent random variables. We explain first how to derive the distribution function of the sum and then how to derive its probability mass function (if the summands are discrete) or its probability density function (if the summands are continuous).

## Distribution function of a sum

The following proposition characterizes the distribution function of the sum in terms of the distribution functions of the two summands.

Proposition Let and be two independent random variables and denote by and their distribution functions. Letand denote the distribution function of by . The following holds:or

Proof

The first formula is derived as follows:The second formula is symmetric to the first.

Example Let be a uniform random variable with support and probability density functionand another uniform random variable, independent of , with support and probability density functionThe distribution function of isThe distribution function of isThere are four cases to consider:

1. If , then

2. If , then

3. If , then

4. If , then

By combining these four possible cases, we obtain

## Probability mass function of a sum

When the two summands are discrete random variables, the probability mass function of their sum can be derived as follows.

Proposition Let and be two independent discrete random variables and denote by and their respective probability mass functions and by and their supports. Letand denote the probability mass function of by . The following holds:or

Proof

The first formula is derived as follows:The second formula is symmetric to the first.

The two summations above are called convolutions (of two probability mass functions).

Example Let be a discrete random variable with support and probability mass functionand another discrete random variable, independent of , with support and probability mass functionDefineIts support is The probability mass function of , evaluated at isEvaluated at , it isEvaluated at , it isTherefore, the probability mass function of is

## Probability density function of a sum

When the two summands are absolutely continuous random variables, the probability density function of their sum can be derived as follows.

Proposition Let and be two independent absolutely continuous random variables and denote by and their respective probability density functions. Let:and denote the probability density function of by . The following holds:or

Proof

The distribution function of a sum of independent variables isDifferentiating both sides and using the fact that the density function is the derivative of the distribution function, we obtainThe second formula is symmetric to the first.

The two integrals above are called convolutions (of two probability density functions).

Example Let be an exponential random variable with support and probability density functionand another exponential random variable, independent of , with support and probability density functionDefine The support of isWhen , the probability density function of isTherefore, the probability density function of is

## More details

### Sum of n independent random variables

We have discussed above how to derive the distribution of the sum of two independent random variables. How do we derive the distribution of the sum of more than two mutually independent random variables? Suppose , , ..., are mutually independent random variables and let be their sum:The distribution of can be derived recursively, using the results for sums of two random variables given above:

1. first, defineand compute the distribution of ;

2. then, defineand compute the distribution of ;

3. and so on, until the distribution of can be computed from

## Solved exercises

Below you can find some exercises with explained solutions.

### Exercise 1

Let be a uniform random variable with support and probability density functionand an exponential random variable, independent of , with support and probability density functionDerive the probability density function of the sum

Solution

The support of isWhen , the probability density function of isTherefore, the probability density function of is

### Exercise 2

Let be a discrete random variable with support and probability mass functionand another discrete random variable, independent of , with support and probability mass functionDerive the probability mass function of the sum

Solution

The support of is The probability mass function of , evaluated at isEvaluated at , it isEvaluated at , it isEvaluated at , it isTherefore, the probability mass function of is

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